Question

The magnitude of vectors $\overrightarrow{A}$, $\overrightarrow{B}$ and $\overrightarrow{C}$ are 3, 4 and 5 units respectively. If $\overrightarrow{A} + \overrightarrow{B} = \overrightarrow{C}$, the angle between $\overrightarrow{A}$ and $\overrightarrow{B}$ is

• A. $\pi / 2$
• B. $\cos^{ - 1} (0 . 6)$
• C. $\tan^{ - 1} \, (7/5)$
• D. $\pi / 4$

Answer 1

The Correct Option is A

Let $\theta$ angle between $\overrightarrow{A}$ and $\overrightarrow{B}$
Given : $A = | \overrightarrow{A} | = 3 \, units$
$B+ | \overrightarrow{B} | = 4\, units$
$C = | \overrightarrow{C} | = 5 \,units$
$\overrightarrow{A} + \overrightarrow{B} + \overrightarrow{C}$
$\therefore ( \overrightarrow{A} + \overrightarrow{B} ) \cdot ( \overrightarrow{A} + \overrightarrow{B} ) = \overrightarrow{C} \cdot \overrightarrow{C}$
$\overrightarrow{A} \cdot \overrightarrow{A} + \overrightarrow{A} \cdot \overrightarrow{B} + \overrightarrow{B} \cdot \overrightarrow{A} + \overrightarrow{B} \cdot \overrightarrow{B} = \overrightarrow{C} \cdot \overrightarrow{C}$
$A^2 + 2 A B \cos \theta + B^2 = C^2$
$9 + 2AB \cos \theta + 16 = 25$ or $2 A B \cos \theta = 0$
or $\cos \theta = 0 \therefore \theta = 90^\circ$.